calculation in C program always results in 0 whenever using division

问题

I'm using two different variable to divide in the calculation with the variable from int and double. These work fine when I use something like:

int cost
cost = 40;
cost = (cost / 400) * 20 * 2;

For this the method works fine and I get the right result which is 4, but when I use the variable cost and put it in the header instead, like:

#define cost 40
int total_cost;
total_cost = (cost / 400) * 20 * 2;

this always results in 0 for me and I don't know why. Even if I use printf with %d or %f this still gives me a result of 0.

回答1:

You are doing integer division - which rounds down.

Therefore:

cost / 400

is returning zero because cost = 40 and 40 / 400 rounds down to zero.

What you should do is use a floating-point type like double.

EDIT:

double cost
cost = 40;
cost = (cost / 400) * 20 * 2;

and

#define cost 40
double total_cost;
total_cost = ((double)cost / 400) * 20 * 2;

回答2:

Order of operations, and Integer Division.

Integer Division always truncates. Mathematically, 40/400 = .1 - but that's not an integer. The remainder is thrown away, leaving you with: 40 / 400 = 0.

Order of operations means that the division is done first, in your example. Since the order of multiplication and division doesn't matter too much (mathematically speaking), try changing the order:

total_cost = cost * 20 * 2 / 400;

Multiplication happens first, division last, giving you:

40 * 20 * 2 / 400 = 800 * 2 / 400 = 1600 / 400 = 4

来源：https://stackoverflow.com/questions/7720078/calculation-in-c-program-always-results-in-0-whenever-using-division

标签

math